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PilotLPTM [1.2K]
3 years ago
7

The rate constant of a reaction is 5.8 × 10−3 s−1 at 25°c, and the activation energy is 33.6 kj/mol. what is k at 75°c? enter yo

ur answer in scientific notation.
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

k₂ = 4.06 x 10⁻² s⁻¹.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₁ = 5.8 × 10⁻³ s⁻¹, k₂ = ??? , Ea = 33600 J/mol, R = 8.314 J/mol.K, T₁ = 298.0 K, T₂ = 348.0 K.

  • ln(k₂/5.8 × 10⁻³ s⁻¹) = (33600 J/mol / 8.314 J/mol.K) [(348.0 K - 298.0 K) / (298.0 K x 348.0 K)] = (4041.37) (4.82 x 10⁻⁴) = 1.9479.
  • Taking exponential of both sides:

(k₂/5.8 × 10⁻³ s⁻¹) = 7.014.

∴ k₂ = 4.06 x 10⁻² s⁻¹.

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What is the molarity of a solution of 58.7 grams of MgCl2 in 359 ml of solution?
jekas [21]

Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

= 24 + 71

= 95g/mol

mole = 58.7g ÷ 95g/mol

mole = 0.618mol

Volume of solution = 359ml = 359/1000 = 0.359L

Molarity = 0.618mol ÷ 0.359L

Molarity = 1.72 M

6 0
3 years ago
The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what pe
Trava [24]

Answer:

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

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Initial concentration:  0.1    →  3.98 ^10-3       +      0

equilibrium concentration:  0.1(1-α) →   3.98 * 10-3 + 0.1α              0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)      

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

6 0
3 years ago
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